Question

A skateboarder travels on a horizontal surface with an initial velocity of 4.0 m/s toward the south and a constant acceleration of 1.8 m/s2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.

a)What is her x position at t=0.60s?

b)What is her y position at t=0.60s?

c)What is her x velocity component at t=0.60s?

d)What is her y velocity component at t=0.60s?

Answer 1

Answer:

a) 0.32 m b) -2.4 m c) 1.08 m/s d) -4 m/s

Explanation:

a)

• As the x and y axes (as chosen) are perpendicular each other, the movements along these axes are independent each other.
• This means that we can use the kinematic equations for displacements along both axes.
• In the x direction, as the only initial velocity is in the south direction (-y axis), the skateboarder is at rest, so we can write:

        $x =\frac{1}{2}*a*t^{2} (1)$

• In the y-direction, as no acceleration is acting on the skateboarder, we can write  the following displacement equation:

        $y = v_{0y} * t (2)$

• For t = 0.6s, replacing by the givens, we get the position (displacement from the origin) on the x-axis, as follows:

       $x =\frac{1}{2}*a*t^{2} =\frac{1}{2} * 1.8 m/s2*(0.6s)^{2}\\ x = 0.32 m$

b)

• From (2) we can get the position on the y-axis (displacement from the origin) as follows:

        $y = v_{0y} * t = -4 m/s * 0.6 s = -2.4 m$

c)

• In the x- direction, we can find the component of the velocity along this direction, as follows:

        $v_{fx} = a*t$

• Replacing by the values, we have:

        $v_{fx} = a*t = 1.8 m/s2 * 0.6 s = 1.08 m/s$

d)

• As the skateboarder moves along the y-axis at a constant speed equal to her initial velocity, we  have:

        vfy = voy = -4 m/s